Suppose you have a tube 1.20 m long, open on both ends. Determine the frequency

Question

Suppose you have a tube 1.20 m long, open on both ends. Determine the frequency of the 2nd overtone, assuming the speed of sound to be 343 m/s. Repeat for the same length tube but with a closed end.

Explanation / Answer

1. Path length difference = 20 cm. pi*r-2r = 20 cm ==> r = 20/(pi-2) cm 2. Ref. (0 db) intensity level I0 = 1E-12 w/m^2. I(70) = I0*10^(dB/10) = 1E-5 w/m^2 I(50) = I0*10^(dB/10) = 1E-7 w/m^2 Amplitude is a bit tricky since it's related to sound pressure (and intensity) by impedance. The impedance used below is 400 Pa-s/m, which is the nominal impedance used to relate ref. pressure to ref. intensity. (However, actual impedance is a complex function of air pressure and density.) Pressures and amplitudes are given below. Ref. (0 db) pressure P0 = 2E-5 Pa. omega = 2pi*f = 3141.59265 rad/s P(70) = P0*10^(dB/20) = 6.3245553E-2 Pa A(70) = P(70)/(omega*Z) = 5.033E-8 m P(50) = P0*10^(dB/20) = 6.3245553E-3 Pa A(50) = P(50)/(omega*Z) = 5.033E-9 m 3. A. Resonant pipe fundamental wavelength lambda = 4L, fundamental frequency f = c/lambda B. Since I don't know the intended value for c, I can't give a numeric answer. However from the basic equation f = sqrt(T/(m/L))/(2L) you can solve for tension T. T = 4f^2*mL (L here is the length of the wire).

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